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Basing on the definition linear algebra gives, a linear real valued function of real values $f$ is a function such that: $$f(x+y)=f(x)+f(y)$$ $$f(ax)=af(x)$$ Is there any function $g: A\subseteq\mathbb{R}\rightarrow\mathbb{R}$, complex as you want or nonelementary, that posses the same properties of a linear function (additivity and homogeneity), but isn't a line passing through the origin?

In other words, is this double implication valid?

$f: B\subseteq\mathbb{R}\rightarrow\mathbb{R}$ is a linear function $\Leftrightarrow$ $\exists C\in\mathbb{R}$ such that $f(x)=Cx$

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    I don't get it. These properties are the definition of a linear function. So by definition any function that possess them aka "has the same caratheristics" is linear. If you doubt only geometrical part that linear functions' graphs can be not straight lines, then, well, what's your definition of a straight line? – 2017-02-10
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    How do you define a straight line? Is $\{(x,y)\in \mathbb{R}^2:x\in \mathbb{Q},x=y\}$ a straight line to you? – 2017-02-10
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    If you're only considering functions from $\mathbb{R}$ to $\mathbb{R}$, then yes: A function $f\colon\mathbb{R} \to \mathbb{R}$ is linear (in the Linear Algebra sense of "linear") if and only if, for some $a \in \mathbb{R}$, $f(x) = ax$ for all $x \in \mathbb{R}$. – 2017-02-10
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    $f(x)=0$ satisfies your equations for a linear function. The way you have formulated this makes it hard to analyse - what are supposed to be the range and domain of your function? In what space does the straight line you imagine live? Are you looking at a real valued function of real numbers and the points $(x,f(x))$ in the real plane? In multiple dimensions, so when $x$ is a vector in three dimensions for example, linear maps can be described by matrix multiplication. – 2017-02-10
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    Well, the graphic of a linear function $f:\Bbb R^3\to \Bbb R^5$ is a linear subspace of dimension $3$ in $\Bbb R^8$. So no "straight line". – 2017-02-10
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    "straight line" is one of the great pleonasms. the adjective serves no function; all lines are straight! we never speak of *unstraight* lines. (we may call them geodesics, but that's another matter...) – 2017-02-10
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    I'm sorry for the ambiguity of the question. I wasn't thinking to multi-dimensional spaces, but to regular 2-D geometry, so functions from R to R. Thus, by "straight line passing through the origin" I meant a function from R to R defined as f(x)=Cx, with C being a constant. I proceeded to edit the post in order to clarify the question. – 2017-02-11

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If you weaken the assumption of linearity to additivity, namely only that it satisfies Cauchy's functional equation for every $x,y$, $$ f(x+y) = f(x) + f(y)$$ then it has "non-line" solutions that are in particular, discontinuous, non-monotonic, unbounded on every interval and not Lebesgue measurable. Wikipedia has some details.

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If you only care about linear functions $f: \mathbb{R} \to \mathbb{R}$, then all linear functions have a graph that is a line. Since $f(ax) = af(x)$, you have that $f(0) = f(0 \cdot 0) = 0f(0) = 0$. Then $$ f(x) = f(x \cdot 1) = xf(1). $$ So, the graph is a line of slope $f(1)$.

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Take $f(ax) = af(x)$. Substitute $x = a^u$ into it, then define a new function $g(u) = f(a^u)$. This allows us to rewrite our function as: $$g(u+1) = ag(u)$$ This can be solved using regular difference equation methods, to find that $g(u) = Ca^u$. This also implies that $f(x) = Cx$ and that is the only function it can be.

Interestingly enough, this makes the first property you listed redundant. If $f(ax) = af(x)$, then $f(x + y) = f(x) + f(y)$.

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    It may be worth an explicit statement that you're restricting to functions $f:\mathbb K\to\mathbb K$ here, where $\mathbb K$ is the scalar field. (I suspect in fact you meant to use $\mathbb K=\mathbb R$.) – 2017-02-10
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    What about taking an arbitrary function $g:[0, 1)\to \mathbb{R}$ and extending it to $\mathbb{R}$ by $g(u + n) = a^n g(u)$ for $u\in [0, 1)$ and $n\in \mathbb{Z}$? – 2017-02-11
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    I think defining the set we're working with is above the scope of this question. It's clear what the OP is working with. Why complicate the question further than it needs to be complicated? – 2017-02-11